How do you reduce amperage?

[AznGothic]

Ok, I got this little project of mine that uses a 6V DC motor. Initially I used 4 D cell batteries which provides 6V and about 4-5amps. The only problem is the D cell batteries don't have a very high maH rating so they die rather quickly which gets pretty costly to constantly buy more D batteries. I tried looking for some D size rechargeable batteries but can't find them anywhere in my area. I finally went to the next city over and found some rechargeable D size but no charger. I finally gave up on the D batteries and bought myself a 6V remote control car battery pack that came with it's own charger. My only problem is, the D batteries were putting out about 4.5amps constant with an initial draw of around 7amps to spin up the motor but the remote control battery is putting out 15amps. When I wire up the R/C car battery to the motor, the motor spins so fast that it starts burning up. How can I reduce the 15amps to around 5-7amps and keep the 6V it's putting out?

BTW, I also tried using a AC to DC converter but the highest one I could find was only 2amps from Radio Shack and that would barely spin the motor at all.

[fmcwilliams]

This is not really my area of expertise so I only offer this as a suggestion, and perhaps one of our other experts can provide a more elegant solution. According to Ohms Law (I=E/R or Amps=Volts/Ohms) if the voltage in the circuit remains constant and the resistance is increased, the current should decrease. Try adding resistance to the circuit to lower the current output.

[AznGothic]

See I was thinking buying a resistor and soldering it between the positive lead of the battery and motor would reduce amperage but I wanted to make sure before I went out and bought a resistor. What made me have doubts of that working was cause I thought with less amperage you have less voltage. Like in a battery, when it starts to die it's because you've used up all the amps and as a result it loses voltage. Atleast that's how it was explained to me. So I was worried putting a resistor in the circuit would also lower my voltage.

Atleast now I know the formula for it. Resistors are pretty cheap so I could use the formula to find out what size resistor I'd need and then try it out. Do you know if I'd have to find out the internal resistance of the battery first and then figure out the resistance I'd need?


I just did a little more research and if I'm understanding all of this correctly, it's the wattage that really matters not the voltage or amperage? I'm getting this from Ohm's Law P=EI. So if my motor is a 6V 7A motor, that means 6V*7A=42W so my motor is using 42watts of power. My RC car battery is 6V 15A (6V*15A=90watts) so it's putting out 90watts of power. So is it safe to say even if I lost some voltage while lowering the amperage, as long as it still equals out to 42watts it'll be fine?

[AznGothic]

Sorry, P=EI is Watt's Law not Ohm's Law....I was reading while typing and ended up typing what I was reading.

[phi-unit]

As far as I know in the net amount of charge will always be conserved. You can't just get amps to disappear but to to reduce the amperage to your motor you wire your resistor into a parallel configuration with your motor. Depending on the resistance of your resistor(since current will flow through the path of least resistance) some of the current from your battery will go to your motor and the other portion to your resistor. If you want to know a little bit more about the parallel configuration I mentioned here is a link:http://en.wikipedia.org/wiki/Parallel_circuit
I hope I have answered your question but if not feel free to post a response.

Phi-unit

[Craig_Bridge]

Ok, I got this little project of mine that uses a 6V DC motor. Initially I used 4 D cell batteries which provides 6V and about 4-5amps. The only problem is the D cell batteries don't have a very high maH rating so they die rather quickly which gets pretty costly to constantly buy more D batteries. I tried looking for some D size rechargeable batteries but can't find them anywhere in my area. I finally went to the next city over and found some rechargeable D size but no charger. I finally gave up on the D batteries and bought myself a 6V remote control car battery pack that came with it's own charger. My only problem is, the D batteries were putting out about 4.5amps constant with an initial draw of around 7amps to spin up the motor but the remote control battery is putting out 15amps. When I wire up the R/C car battery to the motor, the motor spins so fast that it starts burning up. How can I reduce the 15amps to around 5-7amps and keep the 6V it's putting out?

The other experts gave you some basic help in terms of ohms law; however, applying that to analysis of your circuit is a bit tricky.

While the open circuit voltage (current draw limited to what a DC volt meter uses) of 4 D cells in series is probably 6 volts, once you start drawing large amounts of current (4-5 Amps) from them, their internal resistance (along with any resistance in your connection wiring) will reduce the voltage across the motor. To fully characterize the operating point of this circuit, you have to measure both the voltage and current simultaneously.

Based on your remote control battery pack use causing the motor to spin faster and overheat, it obviously has a lower internal resistance and is capable of supplying more current to your motor.

Do you have a data sheet for your DC motor? Data sheets for DC motors will typically provide some curves that show voltage and current relationships for various speed and torque output. Motors also have some duty cycle ratings. In other words, they may or may not be designed to run continuously.

If you have no mechanical load on the motor (shaft is free and unattached) and a high current 6VDC supply will overspeed and overheat your motor, it probably isn't designed to run on 6VDC. I say this because the no mechanical load operating point is usually the lowest current draw operating point for a given voltage on any DC motor. Any mechanical load will increase the current requirements with the maximum curernt occuring at a locked rotor state (more mechanical load than what the motor can turn).

Simple DC motor control circuits typically consist of high power variable resistors in series between the power source and the motor. Determining the appropriate power rating and resistance range requires a lot more knowledge of the internal resistance of your power source and the DC motor properties than what you have shared.

What are you using this DC motor for? What application was your DC motor originally designed for? Knowing the design and usage applications may help us guide you in useful directions.

[AznGothic]

Sad to say I don't have any data sheets on the motor. It was originally used for a portable air pump which actually took 4 D batteries. I'm not sure if you know what the internals of a turbo charger for a car looks like but that's basically how this pump was configured. I know the current the motor will draw ranges from free load being the least amount of current to stall being the highest amount of current. Seeing how I don't have a data sheet for the motor since it came out of an air pump, I don't know what these ranges are so I figured I'd just go by what it was originally designed to use, the 4 D cell batteries, and test the power ratings that way. I'm using the motor to demonstrate how the angle of the blades on a fan can change air speed and quantity (cfm). Or atleast that's what I'm hoping to demonstrate. The pump the motor came out of is actually not that big and the "fan" that it spun was pretty lite weight so taking into consideration that it had to suck air in one port and blow air out the other, I'm guessing it wasn't to far off from free load.

If I'm understanding what you said correctly, if the motor is indeed designed to run 6VDC then no matter how high of amperage my power source is, the motor will never draw more then it needs to run at maximum speed (free load speed)?

BTW, I wanted to thank those of you who have responded to my question and apologize if I seem to be ungrateful or unproductive since I haven't really acted upon any of the advice given to me yet. I've just been extremely busy lately trying to pick a college and a degree program but I am still very interested in my project and any advice/help that I can obtain.

[Craig_Bridge]

It was originally used for a portable air pump which actually took 4 D batteries.

If the original application was to blow up air mattresses, then that gives me a few clues about the motor. 1) It is probably not a 100% duty rated design. It probably came with warnings about not using it for too long at one time to prevent over heating. 2) It was probably designed to operate with forced air cooling. Part of its energy was used to move cooling air over the motor itself to keep it cool. 3) It probably relied on the internal resistance of the 4 D cells and their discharge characteristics as an integral part of the current/power limiting aspects of the design.

The pump the motor came out of is actually not that big and the "fan" that it spun was pretty lite weight so taking into consideration that it had to suck air in one port and blow air out the other, I'm guessing it wasn't to far off from free load.

A rough guess at the mechanical load on the motor (useful delivered work per minute) for a fan can be calculated by the cubic feet per minute of air that is being moved and the pressure difference. My guess would be that the amount of work done by moving air "open circuit" (e.g. not connected to an air mattress so it isn't working against any back pressure of an air mattress) makes its "minimum load operating point" considerably different than the free (or no) load operating point. The compound fan design including the motor cooling air flow is probably acting as part of a speed limiter. The faster the motor runs, the more back pressure the fan blades incur so the more torque the motor must produce which means the more current is required; however, the 4 D cell batteries can only produce so much current because of their internal resistance. In other words, this "simple" design is really a complex and sophisticated feedback mechanism.

If I'm understanding what you said correctly, if the motor is indeed designed to run 6VDC then no matter how high of amperage my power source is, the motor will never draw more then it needs to run at maximum speed (free load speed)?

Yes, this is how self limited 100% duty rated DC motors voltage ratings work for ambient temperatures where simple thermal conductivity (aka. self cooling) works. Unfortunately, I don't think you are working with a DC motor that is 100% duty rated at 6 VDC without cooling.

I'm using the motor to demonstrate how the angle of the blades on a fan can change air speed and quantity (cfm). Or at least that's what I'm hoping to demonstrate.

1) Do you have a way to measure both the voltage across the DC motor and the current through the motor and the RPM concurrently? IMO: This is what it is going to take to characterize the motor's operating curve.
2) How were you planning on measuring CFM?
3) Do you have a way to force air cool the motor using a separate fan without affecting airflow of your test setup?
4a) Do you have access to some high power low resistance resistors? The power factor for resistance heat dissapation is current squared times resistance. If you want to limit the locked rotor current to 6 amps from your 6 volt supply you will need a 36 W 1 ohm resistor. Because you are going to need a variety of resistance values, you probably are going to want to come up with ten 0.1 ohm 5 Watt resistors that you can connect in series, or 4b) Do you have access to a low voltage high current lab DC power supply? Something that can be adjusted from 0 to 6 VDC and current limited from .1 Amp to 6 Amps?

BTW, I wanted to thank those of you who have responded to my question and apologize if I seem to be ungrateful or unproductive since I haven't really acted upon any of the advice given to me yet. I've just been extremely busy lately trying to pick a college and a degree program but I am still very interested in my project and any advice/help that I can obtain.

Thanks is appreciated but no appologies are needed! You are the investigator. Feel free to proceed at your own pace and even ignore our help. It is your project and your learning experience and your life. We are just here to assist you if we can in your science adventure.

[AznGothic]

1) Do you have a way to measure both the voltage across the DC motor and the current through the motor and the RPM concurrently? IMO: This is what it is going to take to characterize the motor's operating curve.
2) How were you planning on measuring CFM?
3) Do you have a way to force air cool the motor using a separate fan without affecting airflow of your test setup?
4a) Do you have access to some high power low resistance resistors? The power factor for resistance heat dissapation is current squared times resistance. If you want to limit the locked rotor current to 6 amps from your 6 volt supply you will need a 36 W 1 ohm resistor. Because you are going to need a variety of resistance values, you probably are going to want to come up with ten 0.1 ohm 5 Watt resistors that you can connect in series, or 4b) Do you have access to a low voltage high current lab DC power supply? Something that can be adjusted from 0 to 6 VDC and current limited from .1 Amp to 6 Amps?

I can get 2 multimeters and set one to voltage and then other to amperage. As far as RPM goes, I don't really have a way to test that. I'm not exactly sure as of yet how I'm going to measure CFM. I have a friend that says for some unknown reason he has a anemometer packed up somewhere. He was going to look for it, dig it out and see if it even still works. If it doesn't then I'll have to go another route to measure the cfm. I was planning all this out and decided to make sure the motor worked first and that's when I ran into problems with the whole battery thing. I was thinking if my friend can't find his or it doesn't work, then I could make-shift a homemade wind tunnel of sorts with a anemometer made from 4 cups (standard weather gadget type thing). If I can count how many rpm's the cups turn and multiply that by the circumfrance that the cups make in feet I should get a rough estimate. At the very least I should be able to prove how fan blade angles change cfm. I'm not sure if I could force air cool the motor using a seperate fan since some people might argue that it would throw my readings off but the motor, when it was in the pump, didn't have any kind of fan cooling system. It was just self cooled or cooled with standing air. As far as resistors goes, I'd have to make a trip to radio shack or order them online (hate online shopping). Also, could you go into a little more detail on how you got 36W 1ohm? I wish I had access to a low voltage high current DC power supply. That would simplify things greatly lol.

By duty cycle do you mean how long the motor can run with no forced air cooling under load and not burn up? I can see what you mean by the simple design being more complex then it seems. I've never even looked at it the way you explained it but when you really think about it, it makes a lot of sense. And after reading all of this plus doing a little more research on my part in my spare time, I don't think this motor was designed to run 6VDC either. Basically because when the D cell was tested it tested at 7-8amps but with the motor connected it dropped to 4-5amps. Correct me if I'm wrong but this means from the internal resistance of the batteries themselves plus the resistance of the motor it dropped amperage which would in turn also drop voltage.

[AznGothic]

hmm....seems I made a mistake. Multiplying rpm's by circumference would give me air velocity not cfm seeing how cfm is more of a measure of air volume. Is there a way for me to get volume from velocity?

[Craig_Bridge]

By duty cycle do you mean how long the motor can run with no forced air cooling under load and not burn up?

Yes and no. Duty cycle for motors are typically specified for some operating temperature and assume that any forced air cooling built into the system is operating.

I don't think this motor was designed to run 6VDC either. Basically because when the D cell was tested it tested at 7-8amps but with the motor connected it dropped to 4-5amps. Correct me if I'm wrong but this means from the internal resistance of the batteries themselves plus the resistance of the motor it dropped amperage which would in turn also drop voltage.

Your very close to a valid interpretation; however, where the voltage drop occurs matters. Your current meter has a shunt resistance, the batteries have a combined internal resistance, and the motor has some voltage/current/speed/torque/temperature plot that is effectively some resistance at any given operating point. Analyzing this, you have a 6 VDC power source flowing current through three series resistsors. The motor's design voltage can be estimated by measuring the voltage directly across the motor when it is operating within its design parameters and duty cycle. This is why you are going to require two meters in the circuit at the same time so that the circuit doesn't change between measuring current and measuring voltage.

Also, could you go into a little more detail on how you got 36W 1ohm?

I assumed that the locked rotor curernt for the motor was infinite (not reality, but a useful simplification) which means it is 0 Ohms. The circuit then being analyzed is simplified to a battery and a this unknown resistor. Since I don't know what the internal resistance of the batteries are, I arbitrarily picked a 1 Ohm resistor (nice simple easy to compute) to see what happened. An infiinite current 6 VDC power supply (again a useful over simplification) will yield 6 Amps of current flow. Current squared time resistance equals watts so a 36 Watt 1 ohm resistor will be within its current limitations. Based on the 4-5 amp operating point data you provided, 1 ohm is probably at the high end of being big enough given that the internal battery resistance and the motor resistance is in series to be a useful current limit and that smaller resistance values are appropriate.
I next arbitrarily divided this 1 ohm resistor into 10 pieces to come up with 0.1 Ohm resistors which need to handle at least 3.6 Watts and I went up to the next standard wattage size of 5 Watts to provide a reasonable margin of safety on thermal disapation. In other words, I oversimplified the problem to come up with something that had a high probability of working that would allow experimenting safely.

Is there a way for me to get volume from velocity?

Maybe. If you can assume a laminar flow (all cross sections of equal area have the same flow velocity and pressure), then the cross sectional area of the flow (square feet) times the air velocity (feet / minute) will give you cubic feet / minute. Knowledge of fan blade (simple propeller) aerodynamics tells us that the flow is not laminar in the plane of the fan blades; however, if the flow is constrained in some pipe, then several feet down the pipe, one would expect a laminar flow to develop. One can decrease the distance required by putting in an equal cross section vane.

[peteryoung]

Hi, let me put in my $.02 cents here. I used to write an electric propulsion column
for an r/c magazine and think I can help clear up some of the issues.

First, the 4 D-cell batteries. D-cell batteries have very high internal impedance which, in plain English, means they are limited in their current output capacity. That makes sense as D-cell batteries are normally used to power low power applications like flashlights, small lightloaded electric motors, etc.
Second, the r/c car battery pack you bought is probably either Nickel Cadmium or Nickel Metal Hydrides; either is rechargeable but the real tipoff is -- these "chemistries" have very LOW internal impedances which means -- they're perfect for putting out mucho current which, when coupled with fairly stout voltage drops, = lots of power available. Which is exactly what r/c car (and r/c airplane) people demand. <As you found out, If you try to demand high currents from the D-cells, the voltage drops and the total capacity diminishes. Contrast that with small "button" cells in your watch or IPod which, as they deliver very low currents, will run for a long time before needing replacement or recharging).
Third, looking at the 7 amps for the D-cells and the 15 amps for the r/c car battery pack: these are VERY HIGH current levels which tells me that your electric motor is under severe load - it must be turning a fairly big "flywheel" or geared transmission to be demanding that much current with either battery set. Electric motors have a very unusual characteristic when presented with a high mechanical load: they demand MORE power from the battery and this "stall torque" condition will persist until the battery dies or something melts - but the torque will increase until the bitter end.
Fourth - as you've noted, your electric motor was not happy when loaded down and powered by the r/c car pack - you mentioned that it was "smoking". That's because
it was stalled and possibly, may have damaged itself with all this electrical energy surging through the brushes and into the armature. If you could disconnect the motor
from its mechanical load and just run it with no load, you'll find that the current draw measured will be a lot less than what you measured.

Back to your original question. One good reason for reducing amperage is so your motor won't damage itself any more! But (and perhaps I missed it), I haven't figured out what you're trying to do with the motor - move a weight, move air, whatever. That will be important as you want the motor to operate at its highest efficiency point, in general, so that the majority of the electrical input is being used to produce power and not be turned into heat - which is what is happening now

There are several ways to control the input power and the robotics people and the radio controlled people have these all figured out. Either a voltage or current regulator will be needed and which one will depend on what you're trying to achieve.

I hope this helps!

Peter Young